Question

Find the ${13}^{th}$ term in the expansion of ${(9x-\frac{1}{3\sqrt{x}})}^{18},x\ne 0$

Answer 1

It is known that$(r+1)$ term $\left(T_{r+1}\right)$ in the binomial expansion of $(a+b{)}^n$ is given by $T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$
Thus ${13}^{th}$ term in the expansion of ${(9x-\frac{1}{3\sqrt{x}})}^{18}$ is
$T_{13}=T_{12+1}={}^{18}{C}_{12}(9x{)}^{18-12}{(-\frac{1}{3\sqrt{x}})}^{12}$
$=(-1{)}^{12}\frac{18!}{12!6!}(9{)}^6(x{)}^6{\left(\frac{1}{3}\right)}^{12}{\left(\frac{1}{\sqrt{x}}\right)}^{12}$
$=\frac{\mathrm{18.17.16.15.14.13.12}!}{12!.6.5.4.3.2}.x^6\left(\frac{1}{x^6}\right){.3}^{12}\left(\frac{1}{3^{12}}\right)$

$=18564$